Difference between revisions of "2005 AMC 10A Problems/Problem 13"

(Problem)
(Solution)
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So <math> 4 < n < 130 </math>.  
 
So <math> 4 < n < 130 </math>.  
  
Therefore the answer is the number of [[positive integer]]s over the interval <math> (4,130) </math> which is <math> 125 \Longrightarrow \mathrm{(E)} </math>.
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Therefore the answer is the number of [[positive integer]]s over the interval <math> (4,130) </math> which is <math>129-5+1 = \boxed{\textbf{(E) }125}</math>
  
 
==See also==
 
==See also==

Revision as of 13:20, 13 December 2021

Problem

How many positive integers $n$ satisfy the following condition:

$(130n)^{50} > n^{100} > 2^{200}\ ?$

$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$

Solution

We're given $(130n)^{50} > n^{100} > 2^{200}$, so

$\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}}$ (because all terms are positive) and thus

$130n > n^2 > 2^4$

$130n > n^2 > 16$

Solving each part separately:

$n^2 > 16 \Longrightarrow n > 4$

$130n > n^2 \Longrightarrow 130 > n$

So $4 < n < 130$.

Therefore the answer is the number of positive integers over the interval $(4,130)$ which is $129-5+1 = \boxed{\textbf{(E) }125}$

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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